Isosceles triangle 9

Given an isosceles triangle ABC where AB= AC. The perimeter is 64cm, and the altitude is 24cm. Find the area of the isosceles triangle.

Correct result:

A =  168 cm2


p=64 cm h=24 cm  p=2a+b  a2=(b/2)2+h2 a2=((p2a)/2)2+h2 a2=(p/2a)2+h2 a2=p2/4pa+a2+h2 pa=p2/4+h2  a=p/4+h2/p=64/4+242/64=25 cm  b=p2 a=642 25=14 cm  p1=2 a+b=2 25+14=64 cm p1=p  A=b h2=14 242=168 cm2

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Showing 2 comments:
No, your solution starts with h=25cm, but the question says that h=24cm.

So, looking at one-half of the isosceles triangle, x2 + 242 = (32-x)2    solves to x=7cm,   which gives the area of the isosceles triangle as 7x24 = 168cm2

Dr Math
thank you, we corrected 25 to 24 as altitude....


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